R: If i is true then assign string to matrix

Question

I can't get a for loop to assign the string 'NA' when none of the other conditions are met.

This is what I've tried...

Height <- c(1.6,3.4,0.42,n/a, 0.5,n/a,1.5,0,n/a,22.0)
Height <- matrix(Height)  

h_cat <- matrix(, nrow = length(Height), ncol = 1)
for (i in 1:length(Height)){
  if (Height[i]==0)
    h_cat[i] <- 'NA'
  if (Height[i]>0 & Height[i]<2)
    print(Height[i])
    h_cat[i] <- '0-2 m'
    #print(h_cat[i])
  if (Height[i]>=2 & Height[i]<5)
    h_cat[i] <- '2-5 m'
  if (Height[i]>=5 & Height[i]<10)
    h_cat[i] <- '5-10 m'
  if (Height[i]>=10)
    h_cat[i] <- '>10 m'
  else
    h_cat[i] <- 'NA'
}

I had a go with is.na() but no luck either.

Update

Sorry that was rushed. Data added.

Please share some data... https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example — s_baldur, Aug 25 '17 at 16:08
`h_cat` is a `matrix`, it has row and column numbers. It should be indexed as `h_cat[i, 1]`. — Rui Barradas, Aug 25 '17 at 16:39
You obviously didn't try that code because `n/a` would trigger an error. — Rich Scriven, Aug 25 '17 at 22:16
This is strange code for R which generally tries to avoid control structures like that, but you might just try `h_cat <- Height` and then `h_cat[Height==0] <- NA` etc though I agree that cut is good or even better maybe use a hash. — Elin, Aug 25 '17 at 22:41

score 2 · Accepted Answer · answered Aug 25 '17 at 22:19

There are two main problems with the current code.

First can be exemplified by NA == 0, which returns NA. Thus you can only use the if function on the non-NA entries: which(!is.na(Height)).
The logic of your if-else construction is not working as what seems to be the intention.

The below I believe achieves your intentions:

# Data needs to be numeric to check with ">"
Height[Height == "n/a", ] <- NA
Height <- as.numeric(Height)

h_cat <- matrix(, nrow = length(Height), ncol = 1)

# Can't have NA in logical tests
non_na_entries <- which(!is.na(Height))
for (i in non_na_entries) {
  if (Height[i] == 0) {
    h_cat[i] <- NA
  } else if (Height[i] > 0 & Height[i] < 2) {
    h_cat[i] <- '0-2 m'
  } else if (Height[i] >= 2 & Height[i] < 5) {
    h_cat[i] <- '2-5 m'
  } else if (Height[i] >= 5 & Height[i] < 10) {
    h_cat[i] <- '5-10 m'
  } else if (Height[i] >= 10) {
    h_cat[i] <- '>10 m'
  } else
    h_cat[i] <- NA
}

    h_cat

     [,1]   
 [1,] "0-2 m"
 [2,] "2-5 m"
 [3,] "0-2 m"
 [4,] NA     
 [5,] "0-2 m"
 [6,] NA     
 [7,] "0-2 m"
 [8,] NA     
 [9,] NA     
[10,] ">10 m"

score 0 · Answer 2 · answered Aug 26 '17 at 05:33

There are several problems with your code. Two of them were addressed by snoram. You need to substitute NA for 'n/a', and you don't need Height to be a matrix. To make the code that follows a complete answer, I'll repeat the NA part.

Height <- c(1.6,3.4,0.42,n/a, 0.5,n/a,1.5,0,n/a,22.0)
Height[Height == "n/a"] <- NA

Now, an alternative to a complicated if/else is, like many have said, cut. Since it returns an object of class factor we'll need to recode its return value. To do that, I'll use a function from package dplyr.

library(dplyr)

h_cat <- cut(Height, c(0, 2, 5, 10, Inf))
h_cat[Height == 0] <- NA
h_cat <- recode_factor(h_cat,
                 '(0,2]' = '0-2 m',
                 '(2,5]' = '2-5 m',
                 '(5,10]' = '5-10 m',
                 '(10,Inf]' = '>10 m')
h_cat <- matrix(as.character(h_cat), ncol = 1)
h_cat

I believe that this is much simpler and readable. Neater. If you ever have to come back to this code in the future, you'll probably find it much easier to maintain.

R: If i is true then assign string to matrix

2 Answers2